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Proof step 10 from proof orderedle-all (previous step)

(a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m ≤ 1 + # y, n ≤ 1 + # y, m ≤ n, ∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ a0 < a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]
Proof step 12:
(a0 ' + y)[n -1] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m ≤ n, n ≤ 1 + # y, m ≤ 1 + # y, ∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ n = 0, a0 < a

Rule argument

spec: list-del name: get-numrec seq: n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1] subst: [n, x, a] → [n, a0 ' + y, a] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ (a ' + x)[0] = a; (s)% ⊦ ¬ a < a; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: n, a0 ' + y, a